Photon Momentum and Radiation Force | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Use photon energy to find photon momentum

For an object with mass, momentum is mass multiplied by velocity. A photon has no rest mass, so photon momentum is instead related to photon energy.

p=\dfrac{E}{c}

Symbols

$p$ = momentum of one photon (N s)
$E$ = energy of one photon (J)
$c$ = speed of electromagnetic radiation in a vacuum (m s⁻¹)

Substitute the wavelength form of photon energy from the previous lesson. The factor c cancels and gives the relation below.

p=\dfrac{h}{\lambda}

Photon momentum points in the direction in which the photon travels. At the same wavelength, every photon has the same momentum. A shorter wavelength gives a larger photon momentum.

Higher photon energy means larger momentum.
Shorter wavelength means larger momentum.
Momentum may be written in N s or kg m s⁻¹. The units are equivalent.

Worked example

Calculate the momentum of one photon

A photon has energy 3.11 × 10⁻¹⁹ J. Calculate its momentum (9702/42/F/M/24 Q7(a)).

\begin{aligned} p&=\dfrac{E}{c}\\ &=\dfrac{3.11\times10^{-19}}{3.00\times10^8}\\ &=1.04\times10^{-27}\ \mathrm{N\,s} \end{aligned}

Answer

p=1.04\times10^{-27}\ \mathrm{N\,s}

Your turnrelationship check

Two photons have wavelengths 400 nm and 800 nm. Which photon has the larger momentum, and by what factor?

Show worked answer

The 400 nm photon has twice the momentum because photon momentum is inversely proportional to wavelength.

Absorption and reflection transfer different momentum

Two hand-drawn metal surfaces compare photon absorption by a dark plate with reflection from a polished plate. — Absorption transfers p to the surface. Reflection reverses the photon momentum, so the surface receives 2p. Force is the rate of this momentum transfer.

The figure uses normal incidence: the beam travels perpendicular to the surface. Force is the rate of change of momentum. The surface gains momentum in the original direction of the beam.

Swipe left or right to read every column →

What happens to one photon	Photon momentum change	Momentum given to surface
Absorbed	from +p to 0, so Δp = −p	+p
Reflected straight back	from +p to −p, so Δp = −2p	+2p
Transmitted without a change	no momentum change	0

For a beam that is completely absorbed, the total photon momentum arriving each second gives the force below.

F=\dfrac{P}{c}

Symbols

$F$ = force on the absorbing surface (N)
$P$ = power of the incident beam (W)
$c$ = speed of electromagnetic radiation in a vacuum (m s⁻¹)

If every photon is reflected straight back, each photon reverses its momentum. The momentum transfer and the force are doubled.

F=\dfrac{2P}{c}\qquad\text{complete reflection}

If a fraction a of the beam power is absorbed and a fraction r is reflected straight back, add the two contributions.

F=(a+2r)\dfrac{P}{c}

This expression assumes normal incidence and no other change in the transmitted part of the beam. If the illuminated area is given, radiation pressure is force per unit area.

p_{\mathrm{rad}}=\dfrac{F}{A}

Show why the photon frequency cancels

A photon carries energy E and momentum E divided by c. The photon rate is power divided by E. Multiplying rate by momentum gives:

F=\left(\dfrac{P}{E}\right)\left(\dfrac{E}{c}\right)=\dfrac{P}{c}

Therefore, beams with the same power and beam area exert the same pressure on the same perfectly reflecting surface even if their photon frequencies are different (9702/41/O/N/23 Q8(c)).

Worked example

A beam is half absorbed and half reflected

A 1.0 × 10⁻² W beam reaches a surface. Half the photons are absorbed and half are reflected. Determine the average force (9702/42/F/M/25 Q8(b)(iii)).

The absorbed half produces a force equal to one half of P divided by c.
The reflected half transfers twice as much momentum, producing a force equal to P divided by c.
Add the two contributions.
$\begin{aligned} F&=\dfrac{P}{2c}+\dfrac{P}{c}\\ &=\dfrac{3P}{2c}\\ &=5.0\times10^{-11}\ \mathrm{N} \end{aligned}$

Answer

F=5.0\times10^{-11}\ \mathrm{N}

Common mistake

Use a factor of 2 only for photons that reverse direction. An absorbed photon transfers p; a reflected photon transfers 2p. Apply each fraction to the beam power before adding the forces.

Your turnmomentum-transfer check

The same beam is first fully absorbed and then fully reflected straight back. Compare the forces on the surface.

Show worked answer

Full reflection produces twice the force because each photon changes momentum from forward to backward and transfers twice its initial momentum.

First decide what happens to the photons. Then calculate the momentum change, not only the initial momentum.

Use photon energy to find photon momentum

For an object with mass, momentum is mass multiplied by velocity. A photon has no rest mass, so photon momentum is instead related to photon energy.

p=\dfrac{E}{c}

Symbols

$p$ = momentum of one photon (N s)
$E$ = energy of one photon (J)
$c$ = speed of electromagnetic radiation in a vacuum (m s⁻¹)

Substitute the wavelength form of photon energy from the previous lesson. The factor c cancels and gives the relation below.

p=\dfrac{h}{\lambda}

Photon momentum points in the direction in which the photon travels. At the same wavelength, every photon has the same momentum. A shorter wavelength gives a larger photon momentum.

Higher photon energy means larger momentum.
Shorter wavelength means larger momentum.
Momentum may be written in N s or kg m s⁻¹. The units are equivalent.

Worked example

Calculate the momentum of one photon

A photon has energy 3.11 × 10⁻¹⁹ J. Calculate its momentum (9702/42/F/M/24 Q7(a)).

\begin{aligned} p&=\dfrac{E}{c}\\ &=\dfrac{3.11\times10^{-19}}{3.00\times10^8}\\ &=1.04\times10^{-27}\ \mathrm{N\,s} \end{aligned}

Answer

p=1.04\times10^{-27}\ \mathrm{N\,s}

Your turnrelationship check

Two photons have wavelengths 400 nm and 800 nm. Which photon has the larger momentum, and by what factor?

Show worked answer

The 400 nm photon has twice the momentum because photon momentum is inversely proportional to wavelength.

Absorption and reflection transfer different momentum

The figure uses normal incidence: the beam travels perpendicular to the surface. Force is the rate of change of momentum. The surface gains momentum in the original direction of the beam.

Swipe left or right to read every column →

What happens to one photon	Photon momentum change	Momentum given to surface
Absorbed	from +p to 0, so Δp = −p	+p
Reflected straight back	from +p to −p, so Δp = −2p	+2p
Transmitted without a change	no momentum change	0

For a beam that is completely absorbed, the total photon momentum arriving each second gives the force below.

F=\dfrac{P}{c}

Symbols

$F$ = force on the absorbing surface (N)
$P$ = power of the incident beam (W)
$c$ = speed of electromagnetic radiation in a vacuum (m s⁻¹)

If every photon is reflected straight back, each photon reverses its momentum. The momentum transfer and the force are doubled.

F=\dfrac{2P}{c}\qquad\text{complete reflection}

If a fraction a of the beam power is absorbed and a fraction r is reflected straight back, add the two contributions.

F=(a+2r)\dfrac{P}{c}

This expression assumes normal incidence and no other change in the transmitted part of the beam. If the illuminated area is given, radiation pressure is force per unit area.

p_{\mathrm{rad}}=\dfrac{F}{A}

Show why the photon frequency cancels

A photon carries energy E and momentum E divided by c. The photon rate is power divided by E. Multiplying rate by momentum gives:

F=\left(\dfrac{P}{E}\right)\left(\dfrac{E}{c}\right)=\dfrac{P}{c}

Therefore, beams with the same power and beam area exert the same pressure on the same perfectly reflecting surface even if their photon frequencies are different (9702/41/O/N/23 Q8(c)).

Worked example

A beam is half absorbed and half reflected

A 1.0 × 10⁻² W beam reaches a surface. Half the photons are absorbed and half are reflected. Determine the average force (9702/42/F/M/25 Q8(b)(iii)).

The absorbed half produces a force equal to one half of P divided by c.
The reflected half transfers twice as much momentum, producing a force equal to P divided by c.
Add the two contributions.
$\begin{aligned} F&=\dfrac{P}{2c}+\dfrac{P}{c}\\ &=\dfrac{3P}{2c}\\ &=5.0\times10^{-11}\ \mathrm{N} \end{aligned}$

Answer

F=5.0\times10^{-11}\ \mathrm{N}

Common mistake

Use a factor of 2 only for photons that reverse direction. An absorbed photon transfers p; a reflected photon transfers 2p. Apply each fraction to the beam power before adding the forces.

Your turnmomentum-transfer check

The same beam is first fully absorbed and then fully reflected straight back. Compare the forces on the surface.

Show worked answer

Full reflection produces twice the force because each photon changes momentum from forward to backward and transfers twice its initial momentum.

First decide what happens to the photons. Then calculate the momentum change, not only the initial momentum.